Suppose $I\neq A\neq -I$, where $I$ is the identity matrix and $A$ is a real $2\times2$ matrix. If $A=A^{-1}$ then what is the trace of $A$? I was thinking of writing $A$ as $\{a,b;c; d\}$ then using $A=A^{-1}$ to equate the positions but the equations I get suggest there is an easier way.

We usually consider the eigenvalues, $\lambda$, given by $\det(A-\lambda E) = 0$. In other words, we usually consider the equation $\lambda^2 - \text{tr}({\bf A})\lambda + \det({\bf A}) = 0$. There is a great result due to Hamilton which says that if we replace the number $\lambda$ in the numerical equation $\det({\bf A}-\lambda {\bf E}) = 0$ by the matrix ${\bf A}$ then we get a matrix equation:

In the first case, we get $\text{tr}(A){\bf A} = 0$. Since $\det({\bf A}) \neq 0$ it follows that $\text{tr}({\bf A}) = 0.$ In the second case, $-\text{tr}(A){\bf A} = -2{\bf E}$. This is only possible if ${\bf A} \propto {\bf E}$. For more detail, notice that:

Let $J = V^{-1}A V$ be a Jordan normal form of $A$. Then $J = \mathbb{diag}(J_{\lambda_1,m_1},...,J_{\lambda_k,m_k}) $, where $J_{\lambda,i}$ is a Jordan block with eigenvalue $\lambda$ and size $i \times i$.

Then from $A^2=I$ we have $J^2 = I$, from which we get $J_{\lambda_l,m_l}^2 = I$ for each block. It follows from this that $\lambda_l^2 = 1$, or equivalently, $\lambda_l = \pm 1$.

A brief computation shows that if $m_l>1$, then $[J_{\lambda_l,m_l}^2]_{12} = 2 \lambda_l \neq 0$, from which it follows that $m_l = 1$ for all $l$, and consequently, $J$ is diagonal with entries $\pm 1$.

Since the trace is invariant to similarity transformations, and the dimension is $2$, there are three possible values corresponding to the values $\{(\pm 1) + (\pm 1) \}$. However the values $\{-2, 2\}$ correspond to the matrices $-I$ and $I$ respectively, eliminating them leaves $0$ as the only possible value.